Crux Mathematicorum IN THIS ISSUE / DANS CE NUMÉRO

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1 Crux Mathematicorum VOLUME 38, NO. 10 DECEMBER / DÉCEMBRE 01 IN THIS ISSUE / DANS CE NUMÉRO 398 Mathematical Mayhem Shawn Godin 398 Mayhem Solutions: M507 M The Contest Corner: No. 10 Shawn Godin 405 The Olympiad Corner: No. 308 Nicolae Strungaru 405 The Olympiad Corner Problems: OC106 OC The Olympiad Corner Solutions: OC46 OC50 41 Book Reviews Amar Sodhi 41 Heavenly Mathematics: The Forgotten Art of Spherical Trigonometry by Glen Van Brummelen 413 Mathematical Excursions to the World s Great Buildings by Alexander J. Hahn 415 Problem Solver s Toolkit: No. 3 Murray S. Klamkin 418 Recurring Crux Configurations 9: J. Chris Fisher 40 Problems: Solutions: YEAR END FINALE 439 Index to Volume 38, 01 Published by Publié par Canadian Mathematical Society Société mathématique du Canada St. Laurent Blvd boul. St. Laurent Ottawa, Ontario, Canada K1G 3V4 Ottawa (Ontario) Canada K1G 3V4 FAX: Téléc : subscriptions@cms.math.ca abonnements@smc.math.ca

2 398/ MAYHEM SOLUTIONS MAYHEM SOLUTIONS M507. Proposed by the Mayhem Staff. A 4 by 4 square grid is formed by removable pegs that are one centimetre apart as shown in the diagram. Elastic bands may be attached to pegs to form squares, two different by squares are shown in the diagram. What is the least number of pegs that must be removed so that no squares can be formed? Solution by Juz an Nari Haifa, student, SMPN 8, Yogyakarta, Indonesia. To find the least number of pegs that must be removed, we will take pegs away as necessary. We must consider squares of side length 3,, 1, and 3. We will begin with a 3 by 3 square by eliminating the bottom left peg. Consequently, no 3 by 3 square can be formed. Next we will remove any by squares. There are four different by squares, but one has already been eliminated. Since none of them share vertices, we need to remove 3 squares. Since we are looking for the smallest number of pegs, we will not remove any from the corners so we are left with the pegs shown in the diagram to the right. At this point there is only one 1 by 1 square left so we will proceed by removing the bottom right peg of that square. Due to the pegs we chose to remove so far, neither of the two 5 by 5 squares can be formed. This leaves us with only the by squares to consider. There is only one by left (marked with the open circles in the diagram) and we can remove any one of its pegs to remove the square. Thus we have removed 6 pegs and no more squares can be formed. Note that we can apply the same method to obtain different answers, but the minimum number of pegs that need to be removed will always be 6. Also solved by FLORENCIO CANO VARGAS, Inca, Spain; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; MUHAMMAD LABIB IRFANUDDIN, student, SMP N 8 YO- GYAKARTA, Indonesia; MISSOURI STATE UNIVERSITY PROBLEM SOLVING GROUP, Springfield, MO, USA; TAUPIEK DIDA PALLEVI, student, SMP N 8 YOGYAKARTA, Indonesia; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; and ARIF SETYAWAN, student, SMP N 8 YOGYAKARTA, Indonesia. One incorrect solution was submitted. M508. Proposed by the Mayhem Staff. In 1770, Joseph Louis Lagrange proved that every non-negative integer can be expressed as the sum of the squares of four integers. For example 6 = and 7 = = = (in the theorem it is acceptable to use 0, or to use a square more than once). Notice that 7 had several different representations. Show that there is a number, not greater than that can be represented as a sum of the squares of four distinct non-negative integers in more than 100 ways. (Note that rearrangements are Crux Mathematicorum, Vol. 38(10), December 01

3 MAYHEM SOLUTIONS / 399 not considered different, so = are the same representation of 30.) Solution by Florencio Cano Vargas, Inca, Spain. We are going to count the number of combinations of squares of four integers, (x i, i = 1,, 3, 4), whose sum does not exceed To get a conservative bound let us consider the case where all four integers are equal, i.e. x 1 = x = x 3 = x 4 x. Then we have, x 1 + x + x 3 + x 4 = 4x 10 6 x 500. For any combination four non-negative integers not greater than 500, the sum of the squares of these four numbers is less than Note that there are some combinations, like = < , with some x i > 500 but they are not included in our counting. Therefore we will limit ourselves to combinations x 1 + x + x 3 + x 4 with 0 x i 500 and x i x j when i j. Since each x i can take 501 values, the number of combinations is given by [Note: If we had allowed combinations with repeated integers, we would have combinations with repetition, i.e. we would have had combinations to deal with]. 501 We then have combinations to assign to at most numbers. This gives an average of combinations per integer, hence there is at least one integer not greater than that can be represented as a sum of four distinct non-negative integer in 594 ways by the pigeonhole principle. Note: Following this method we can easily prove that we can find a number not greater than N that can be represented as a sum of squares of four integers in more than 100 ways, where N is the minimal solution of the inequality jè N 4 4 k + 1 N 100 which yields N = It should be noted that one could find a better bound with a more refined š counting method since we have discarded integers x i which are larger than N. Also solved by RICHARD I. HESS, Rancho Palos Verdes, CA, USA; and MISSOURI STATE UNIVERSITY PROBLEM SOLVING GROUP, Springfield, MO, USA. One incorrect solution was submitted. This problem was inspired by an example on page of the book Numbers: A Very Short Introduction by Peter M. Higgins, Oxford University Press, 011. In the example the author showed that there is a number less than that can be written as the sum of four different cubes in 10 distinct ways. M509. Proposed by Titu Zvonaru, Cománeşti, Romania. Let ABC be a triangle with angles B and C acute. Let D be the foot of the altitude from vertex A. Let E be the point on AC such that DE AC and let M be the midpoint of DE. Show that if AM BE, then ABC is isosceles. Copyright c Canadian Mathematical Society, 013

4 400/ MAYHEM SOLUTIONS Solution by Corneliu Mănescu-Avram, Transportation High School, Ploieşti, Romania. We will begin by setting up a set of coordinates such that the side BC is on the x-axis and the altitude AD is on the y-axis. Then we have A(0, a), B( b, 0), C(c, 0) and D(0, 0), where a, b, c (0, ). The equation of line AC is y = a c x + a and since DE AC, the equation of the line DE is y = c ax. Their point of intersection is E a c a +c, ac a +c Š. The midpoint of DE is M and BE are respectively m AM = y M y A = a + c x M x A ac a c (a +c ), ac (a +c )Š. The slopes of the lines AM, m BE = y E y B x E x B = ac a c + b(a + c ). These lines are perpendicular if and only if their slopes satisfy m AM m BE = 1. Thus c(a + c ) = b(a + c ), and since a, b, c > 0, we get b = c, hence it follows that triangle ABC is isosceles. Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; ADRIAN NACO, Polytechnic University of Tirana, Albania; RICARD PEIRÓ I ESTRUCH, IES Abastos, Valencia, Spain; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; MIHAÏ STOËNESCU, Bischwiller, France; KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA; and the proposer. One incorrect solution was submitted. M510. Proposed by Šefket Arslanagić, University of Sarajevo, Sarajevo, Bosnia and Herzegovina. If a, b, c C such that a = b = c = r > 0 and a + b + c 0, compute the value of the expression in terms of r. ab + bc + ca a + b + c Solution by Adrian Naco, Polytechnic University of Tirana, Albania. Based on the properties of the complex numbers, we have that 0 a + b + c = (a + b + c)(a + b + c) = aa + bb + cc + (ab + ab) + (bc + bc) + (ca + ca) = a + b + c + (ab + ab) + (bc + bc) + (ca + ca) = 3r + (ab + ab) + (bc + bc) + (ca + ca). (1) Crux Mathematicorum, Vol. 38(10), December 01

5 MAYHEM SOLUTIONS / 401 Furthermore, ab + bc + ca = (ab + bc + ca)(ab + bc + ca) From (1) and () we get = abab + bcbc + caca + (abbc + abbc) + (bcca + bcca) + (caab + caab) = a b + b c + c a + b (ac + ac) + c (ba + ba) + a (cb + cb) = 3r 4 + r (ca + ca) + r (ba + ba) + r (cb + cb) = r [3r + (ca + ca) + (ba + ba) + (cb + cb)]. () ab + bc + ca = r a + b + c ab + bc + ca = r a + b + c ab + bc + ca a + b + c = r. Also solved by FLORENCIO CANO VARGAS, Inca, Spain; IOAN VIOREL CODREANU, Secondary School student, Satulung, Maramureş, Romania; MARIAN DINCĂ, Bucharest, Romania; GLORIA (YULIANG) FANG, University of Toronto Schools, Toronto, ON; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; ANDHIKA GILANG, student, SMPN 8, Yogyakarta, Indonesia; THARIQ SURYA GUMELAR, student, SMPN 8, Yogyakarta, Indonesia; CORNELIU MĂNESCU-AVRAM, Transportation High School, Ploieşti, Romania; NORVALD MIDTTUN, Royal Norwegian Naval Academy, Sjøkrigsskolen, Bergen, Norway; MISSOURI STATE UNIVERSITY PROBLEM SOLVING GROUP, Springfield, MO, USA; RICARD PEIRÓ I ESTRUCH, IES Abastos, Valencia, Spain; CAO MINH QUANG, Nguyen Binh Khiem High School, Vinh Long, Vietnam; HENRY RICARDO, Tappan, NY, USA; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; and the proposer. M511. Proposed by Gili Rusak, student, Shaker High School, Latham, NY, USA. Pens come in boxes of 48 and 61. What is the smallest number of pens that can be bought in two ways if you must buy at least one box of each type? Solution by Bruno Salgueiro Fanego, Viveiro, Spain. Suppose we buy a boxes with 48 pens each and b boxes with 61 pens each, and in another way we buy c boxes with 48 pens each and d boxes with 61 pens each. Then we have 48a + 61b in one way and 48c + 61d in the other way. Since we want the number of pens to be the same in both ways, we have 48a + 61b = 48c + 61d. If a = c then b = d, in which case the two ways are the same. Since we set up two different ways of buying the pens, then we have a c. Since 48 and 61 are relatively prime, then from 48(a c) = 61(b d) it follows that 61 divides 48(a c). Without loss of generality, we will suppose that a > c. It follows that the minimum possible value of a c is 61, and hence 48(c + 61) + 61b = 48c + 61d, or equivalently, d = 48 + b. Since we must buy at least one box of each type, the smallest possible value of b is b = 1 and then d = 49. Similarly, since the smallest possible value of c is Copyright c Canadian Mathematical Society, 013

6 40/ MAYHEM SOLUTIONS c = 1, it follows that a = 6. Therefore, the smallest number of pens that can be bought is (48)(6) + (61)(1) = (48)(1) + (61)(49) = Also solved by FLORENCIO CANO VARGAS, Inca, Spain; GLORIA (YULIANG) FANG, University of Toronto Schools, Toronto, ON; GESINE GEUPEL, student, Max Ernst Gymnasium, Brühl, NRW, Germany; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; NORVALD MIDTTUN, Royal Norwegian Naval Academy, Sjøkrigsskolen, Bergen, Norway; MISSOURI STATE UNIVERSITY PROBLEM SOLVING GROUP, Springfield, MO, USA; RICARD PEIRÓ I ESTRUCH, IES Abastos, Valencia, Spain; and the proposer. M51. Selected from a mathematics competition. A class of 0 students was given a three question quiz. Let x represent the number of students that answered the first question correctly. Similarly, let y and z represent the number of students that answered the second and the third questions correctly, respectively. If x y z and x + y + z 40, determine the smallest possible number of students who could have answered all three questions correctly in terms of x, y and z. (This questions was originally question 3c from the 008 Galois Contest.) Solution by Florencio Cano Vargas, Inca, Spain. The smallest number of students who have answered all three questions correctly corresponds to a situation where the number of students who have failed at least one question is maximum. Taking into account that the number of students that failed question 1, and 3 are 0 x, 0 y and 0 z respectively, then the number of students who answered at least one question incorrectly is at most: max(0, 0 x + 0 y + 0 z) = max(0, 60 x y z). Since x + y + z 40 then 60 x y z = 60 (x + y + z) = 0. And therefore the number of students who have failed at least one question is at most 60 (x + y + z) 0. The number of students that answered all three questions correctly is at least: 0 (60 x y z) = x + y + z 40. Furthermore, since x + y + z 60 we have the boundaries for the answer as: 0 x + y + z Also solved by RICHARD I. HESS, Rancho Palos Verdes, CA, USA; and BRUNO SALGUEIRO FANEGO, Viveiro, Spain. Crux Mathematicorum, Vol. 38(10), December 01

7 THE CONTEST CORNER / 403 THE CONTEST CORNER No. 10 Shawn Godin The Contest Corner est une nouvelle rubrique offerte pas Crux Mathematicorum, comblant ainsi le vide suite à la mutation en 013 de Mathematical Mayhem et Skoliad vers une nouvelle revue en ligne. Il s agira d un amalgame de Skoliad, The Olympiad Corner et l ancien Academy Corner d il y a plusieurs années. Les problèmes en vedette seront tirés de concours destinés aux écoles secondaires et au premier cycle universitaire ; les lecteurs seront invités à soumettre leurs solutions ; ces solutions commenceront à paraître au prochain numéro. Les solutions peuvent être envoyées à : Shawn Godin, Cairine Wilson S.S., 975 Orleans Blvd., Orleans, ON, CANADA, K1C Z5 ou par couriel à crux-contest@cms.math.ca. Toutes solutions aux problèmes dans ce numéro doivent nous parvenir au plus tard le 1 avril 014. Chaque problème sera publié dans les deux langues officielles du Canada (anglais et français). Dans les numéros 1, 3, 5, 7 et 9, l anglais précédera le français, et dans les numéros, 4, 6, 8 et 10, le français précédera l anglais. Dans la section des solutions, le problème sera publié dans la langue de la principale solution présentée. La rédaction souhaite remercier André Ladouceur, Ottawa, ON, d avoir traduit les problèmes. CC46. Si on place l entrée (m, n) dans la machine A, on obtient la sortie (n, m). Si on place l entrée (m, n) dans la machine B, on obtient la sortie (m + 3n, n). Si on place l entrée (m, n) dans la machine C, on obtient la sortie (m n, n). Nathalie choisit le couple (0, 1) et le place comme entrée dans une des machines. Elle prend ensuite la sortie et la place comme entrée dans n importe quelle des machines. Elle continue de la sorte en prenant la sortie à chaque fois et en la plaçant comme entrée dans n importe quelle des machines. (Par exemple, elle peut commencer par le couple (0, 1) et utiliser successivement les machines B, B, A, C et B pour obtenir la sortie finale (7, 6).) Est-il possible de commencer par le couple (0, 1) et d obtenir la sortie (013013, ) en utilisant les machines dans n importe quel ordre n importe quel nombre de fois? CC47. Un cercle de rayon est tangent aux deux côtés d un angle. Un cercle de rayon 3 est tangent au premier cercle et aux deux côtés de l angle. Un troisième cercle est tangent au deuxième cercle et aux deux côtés de l angle. Déterminer le rayon du troisième cercle. CC48. Déterminer s il existe deux nombres réels a et b de manière que les deux polynômes (x a) 3 + (x b) + x et (x b) 3 + (x a) + x n admettent que des zéros réels. Copyright c Canadian Mathematical Society, 013

8 404/ THE CONTEST CORNER CC49. On a placé une pièce de monnaie sur certaines des 100 cases d un quadrillage Chaque case est à côté d une case case qui contient une pièce de monnaie. Déterminer le nombre minimum de pièces de monnaie qui ont pu être placées. (On dit que deux cases distinctes sont à côté l une de l autre si elles ont un côté commun.) CC50. On considère des entiers positifs de cinq chiffres ou moins. Démontrer que la racine carrée d un tel entier ne peut avoir une approximation décimale qui commence par 0, 1111, mais qu il existe un entier de huit chiffres dont l approximation décimale commence par 0, CC Starting with the input (m, n), Machine A gives the output (n, m). Starting with the input (m, n), Machine B gives the output (m + 3n, n). Starting with the input (m, n), Machine C gives the output (m n, n). Natalie starts with the pair (0, 1) and inputs it into one of the machines. She takes the output and inputs it into any one of the machines. She continues to take the output that she receives and inputs it into any one of the machines. (For example, starting with (0, 1), she could use machines B, B, A, C, B in that order to obtain the output (7, 6).) Is it possible for her to obtain (013013, ) after repeating this process any number of times? CC47. A circle of radius is tangent to both sides of an angle. A circle of radius 3 is tangent to the first circle and both sides of the angle. A third circle is tangent to the second circle and both sides of the angle. Find the radius of the third circle. CC48. Determine whether there exist two real numbers a and b such that both (x a) 3 + (x b) + x and (x b) 3 + (x a) + x contain only real roots. CC49. Coins are placed on some of the 100 squares in a grid. Every square is next to another square with a coin. Find the minimum possible number of coins. (We say that two squares are next to each other when they share a common edge but are not equal). CC50. Show that the square root of a natural number of five or fewer digits never has a decimal part starting , but that there is an eight-digit number with this property. Crux Mathematicorum, Vol. 38(10), December 01

9 THE OLYMPIAD CORNER / 405 THE OLYMPIAD CORNER No. 308 Nicolae Strungaru Toutes solutions aux problèmes dans ce numéro doivent nous parvenir au plus tard le 1 avril 014. Chaque problème sera publié dans les deux langues officielles du Canada (anglais et français). Dans les numéros 1, 3, 5, 7 et 9, l anglais précédera le français, et dans les numéros, 4, 6, 8 et 10, le français précédera l anglais. Dans la section des solutions, le problème sera publié dans la langue de la principale solution présentée. La rédaction souhaite remercier Rolland Gaudet, de l Université de Saint-Boniface, d avoir traduit les problèmes. OC106. Déterminer tous les entiers positifs n pour lesquels tous les entiers à n positions décimales, contenant n 1 uns et 1 seul sept, sont premiers. OC107. Le périmètre du triangle ABC est égal à 4. Des points X et Y sont placés sur les arcs AB et AC de façon à ce que AX = AY = 1. Les segments BC et XY s entrecoupent au point M, à leur intérieur. Démontrer que le périmètre de l un des triangles ABM et AMC est égal à. OC108. Déterminer toutes les fonctions f : R R telles que pour tout x R, y [0, ). f(x) = f(x + y) + f(x + y) OC109. Soit a 1, a,..., a n,... une permutation des entiers positifs. Démontrer qu il existe infiniment d entiers positifs i tels que pgcd(a i, a i+1 ) 3 4 i. OC110. Soit G un graphe qui ne contient pas K 4 comme sous graphe. Si le nombre de sommets de G est 3k, où k est entier, quel est le nombre maximal de triangles dans G? OC Find all the positive integers n for which all the n digit integers containing n 1 ones and 1 seven are prime. OC107. The perimeter of triangle ABC is equal to 4. Points X and Y are marked on the rays AB and AC in such a way that AX = AY = 1. The segments BC and XY intersect at point M in their interior. Prove that the perimeter of one of the triangles ABM or AMC is equal to. Copyright c Canadian Mathematical Society, 013

10 406/ THE OLYMPIAD CORNER OC108. Determine all functions f : R R such that for all x R, y [0, ). f(x) = f(x + y) + f(x + y) OC109. Let a 1, a,..., a n,... be a permutation of the set of positive integers. Prove that there exist infinitely many positive integers i such that gcd(a i, a i+1 ) 3 4 i. OC110. Let G be a graph, not containing K 4 as a subgraph. If the number of vertices of G is 3k, for some integer k, what is the maximum number of triangles in G? OLYMPIAD SOLUTIONS OC46. Let p be a prime number, and let x, y, z be integers so that 0 < x < y < z < p. Suppose that x 3, y 3 and z 3 have the same remainders when divided by p. Prove that x + y + z is divisible by x + y + z. (Originally question 5 from the 009 Singapore Mathematical Olympiad, open section, round.) Solved by Michel Bataille, Rouen, France ; Chip Curtis, Missouri Southern State University, Joplin, MO, USA ; Oliver Geupel, Brühl, NRW, Germany ; David E. Manes, SUNY at Oneonta, Oneonta, NY, USA and Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA. We give the similar solutions of Bataille and Manes. as Let x 3 y 3 z 3 a (mod p). Then the polynomial P (w) = w 3 a factors w 3 a (w x)(w y)(w z) w 3 (x+y+z)w +(xy+xz+yz)w xyz (mod p). Thus Also x + y + z xy + xz + yz 0 (mod p). x + y + z (x + y + z) (xy + xz + yz) 0 (mod p). As 0 < x+y +z < 3p we have x+y +z = p or x+y +z = p. Then if x+y +z = p the claim is obvious, while if x + y + z = p it follows that x + y + z is also even, and hence divisible by p = x + y + z. OC47. Let a, b be two distinct odd positive integers. Let a n be the sequence defined as a 1 = a ; a = b ; a n = the largest odd divisor of a n 1 +a n. Prove that there exists a natural number N so that, for all n N we have a n = gcd(a, b). (Originally question 7 from the 009 India IMO selection test.) Crux Mathematicorum, Vol. 38(10), December 01

11 THE OLYMPIAD CORNER / 407 Titu Zvonaru, Cománeşti, Ro- Solved by Arkady Alt, San Jose, CA, USA and mania. We give the solution of Zvonaru. Let d = gcd(a, b). Since a n 1 + a n = β a n, for some β 1, we can easily deduce that d a n for all n by induction. By induction it is also easy to show that gcd(a n, a n 1 ) = gcd(a, b) = d. Let m = max{a, b}. Then for each n, as a n 1 + a n is even we have a n a n 1 + a n It follows immediately by induction that a n m for all n. Claim: If a k < a k+1 for some k then a n < a k+1 for all n > k + 1. We prove this by induction. Suppose that a k < a k+1 for some k then Similarly, a k+ a k+1 + a k a k+3 a k+ + a k+1 < a k+1 + a k+1 < a k+1 + a k+1. = a k+1. = a k+1, so the claim is true for n = k + and n = k + 3. Suppose the claim is true for n = j and n = j + 1 for some j k +, then a j+ a j+1 + a j < a k+1 + a k+1 = a k+1. Thus the claim is true for all n > k + 1. It follows from here that there can only be finitely many k for which a k < a k+1. Indeed, assume by contradiction that we can find infinitely many i 1 < i < < i n < so that a ij < a ij+1. Then, as i j > i j + 1 it follows from the claim that a ij+1+1 < a ij+1 and hence a i1+1 > a i+1 > > a ij+1 > is a strictly decreasing infinite sequence of positive integers, contradiction. Thus, there are only finitely many k for which a k < a k+1. Hence, there exists a largest such k. Thus, if k 0 denotes the largest such k, we have a n a n+1 for all n k As a n is decreasing from n = k, and as it is positive, it cannot be strictly decreasing. Thus, there exists an m k so that a m = a m+1. As gcd(a m, a m+1 ) = d we get that a m = a m+1 = d, and then it is easy to prove by induction that a n = d for all n m. Copyright c Canadian Mathematical Society, 013

12 408/ THE OLYMPIAD CORNER OC48. The angles of a triangle ABC are π 7, π 7 and 4π 7. The bisectors meet the opposite sides at A, B and C. Prove the A B C is an isosceles triangle. (Originally question from the 009 Columbia Mathematical Olympiad.) Solved by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain ; Michel Bataille, Rouen, France ; Chip Curtis, Missouri Southern State University, Joplin, MO, USA and Titu Zvonaru, Cománeşti, Romania. We give the solution of Bataille. We suppose BAC = 4π 7, ABC = π 7, BCA = π 7 and we denote by a, b, c the side lengths BC, CA, AB respectively. We show that A B = A C. The law of sines yields Thus Similarly thus a sin 4π 7 π cos 7 Š = b sin π 7 Š = = sin 4π 7 sin π 7 π cos = 7 b c. From the bisector theorem we have BC a = C A b BC = Š c Š. π sin 7 Š = a b. = c a + b ; BA = CA = a c b b + c ; AB = B C c a = b c + a, ca a + b ; BA = ca b + c ; CA = ab b + c ; CB = ab c + a. Let I be the incentre of ABC. Then As ABA = π 7 BB = B C. Thus and adding these we get AI AA = b + c a + b + c ; BI BB = a + c a + b + c. = AB B we get IA = IB. Similarly AA = A B and BI = IB = IA = a + c a + b + c B C = ab a + b + c b + c a + b + c AA = c a a + b + c Hence ab a + c = B C = BB = BI + IB a(b + c) = a + b + c. b(a + b + c) = (a + c)(b + c) b = ac + c. Crux Mathematicorum, Vol. 38(10), December 01

13 THE OLYMPIAD CORNER / 409 Also, cos( π 7 ) = cos ( π 7 ) 1 yields a b = b c 1 b3 = ac + bc. Combining the last two relations we get Then, we get and hence Then, ab = ac + bc. BC = c b ; BA = bc a ; CA = b a ; CB = ac b A C = BA + BC BA BC cos π 7 = b c a + c4 b c3 b, A B = CA + CB CA CB cos π 7 = b4 b a c b + b = b4 a + c4 b c. A C A B = c ab (b3 abc + ac ) = c ab (b3 (b 3 bc ) + ac ) = c ab (bc b 3 + ac ) = 0. Thus A B = A C and we are done. [Ed.: Covas mentioned that a proof can be found in Leon Bankoff and Jack Garfunkel, The Heptagonal Triangle, Mathematics Magazine 46 (1973), p. 17.] OC49. Let N be a positive integer. How many non-congruent triangles are there, whose vertices lie on the vertices of a regular 6N-gon? (Originally question 11 from the 009 India IMO selection test.) Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA. We claim there are 3N non-congruent triangles. Draw the regular polygon on a circle of radius R so that the arc length between two consecutive vertices is 1. Then the circle has length 6N. It is easy to see that two triangles with the vertices on the 6N points are congruent if and only if they have the same corresponding arc lengths on the circle. Thus, the number of non-congruent triangles is equal to the number of unordered partitions of 6N into 3 (not necessarily distinct) summands. Thus, denoting by x, y, z the arc lengths, without loss of generality we have Copyright c Canadian Mathematical Society, 013

14 410/ THE OLYMPIAD CORNER x y z and x + y + z = 6N. Thus, the number g(n) of triangles becomes g(n) = #{(x, y) N N x y, x + y 6N y} = #{(x, y) N N x y, x + y 6N} NX 6N x ž = (x 1) x=1 N 1 X = 3N x + 1 (x 1) + = = x = 1 x is odd NX k=1 NX k=1 = 6N 6 (3N k (k )) + (6N 6k + 3) N(N + 1) NX k=1 NX x = 1 x is even (3N k (k 1)) 3N x (x 1) + 3N = 6N 3N 3N + 3N = 3N 6N = (6N 1)(6N ) [Ed.: The stars and stripes method tell us that there are triples of (x, y, z) satisfying the equation. From those, 1 has x = y = z and 3N satisfy x = y z (the cases x = y = 3N, z = 0 and x = y = z = N need to be eliminated). Thus by permutations, 9N 6 have exactly two values equal. The remaining (6N 1)(6N ) 9N have three distinct values. Counting now the unordered triples, the one with (x = y = z) is one, the 9N 6 come in groups of three up to permutations, and the remaining (6N 1)(6N ) 9N come in groups of 6. Thus we have (6N 1)(6N ) 9N N = 36N 18N + 18N N = 3N 3N N 1 = 3N.] OC50. Let n. If n divides 3 n + 4 n, prove that 7 divides n. (Originally question 8 from the 009 India IMO selection test.) Solved by David E. Manes, SUNY at Oneonta, Oneonta, NY, USA. As 3 n + 4 n is odd, it follows that n must be odd. Moreover, as 3 n + 4 n 1 (mod 3) it follows that 3 n. Thus n is an odd integer greater than 3. Let p be the smallest prime dividing n. Let a (mod p). Crux Mathematicorum, Vol. 38(10), December 01

15 THE OLYMPIAD CORNER / 411 Then As 3 is invertible mod p, we get As n is odd, this implies 0 3 n + 4 n 3 n (1 + a n ) (mod p). a n 1 (mod p). ( a) n 1 (mod p). Let r be the order of a modulo p. Then r n and r p 1. As p is the smallest divisor of n, r < p and r n, it follows that r = 1. Hence and hence Thus p 7 and hence p = 7. 1 a (mod p) 3 4 (mod p). Unsolved Crux Problem As remarked in the problem section, no problem is ever closed. We always accept new solutions and generalizations to past problems. Recently, Chris Fisher published a list of unsolved problems from Crux [010 : 545, 547]. Below is one of these unsolved problems. Note that the solution to part (a) has been published [005 : ] but (b) remains open [004 : 49, 43; 005 : ] Proposed by Vasile Cîrtoaje, University of Ploiesti, Romania. Let a 1, a,..., a n be positive real numbers, let r = n a 1 a a n, and let E n = (a) Prove that E n 0 for (a 1 ) n = 3; (a ) n = 4 and r 1; 1 a 1 (1 + a ) + 1 a (1 + a 3 ) a n (1 + a 1 ) n r(1 + r). (a 3 ) n = 5 and 1 r ; (a 4 ) n = 6 and r = 1. (b) Prove or disprove that E n 0 for (b 1 ) n = 5 and r > 0; (b ) n = 6 and r 1. Copyright c Canadian Mathematical Society, 013

16 41/ BOOK REVIEWS BOOK REVIEWS Amar Sodhi Heavenly Mathematics: The Forgotten Art of Spherical Trigonometry by Glen Van Brummelen Princeton University Press, 01 ISBN: , Hardcover, 19 + xvi pages, US$35 Reviewed by J. Chris Fisher, University of Regina, Regina, SK The amusing title, playing on the two meanings of the word heavenly, sets the tone of the book the author with his relaxed and playful style has produced a work that is as enjoyable as it is engrossing. Yet the book is quite serious in the goal suggested by its subtitle, The Forgotten Art of Spherical Trigonometry. With applications to navigation, astronomy, and geography, spherical trigonometry had been an essential part of every mathematician s education until the mid 1950s when the subject suddenly disappeared from North American schools. The thoroughness of its disappearance is evident in the pages of Crux, where geometry problems rarely deal with more than two dimensions, and when they do, they fail to attract much attention. The book is not intended to be a thorough treatment of spherical trigonometry (among other things, the theorems are not presented in their full generality), nor is it a scholarly history of the subject. Instead it is a pleasant blend of history and mathematics. Each topic is introduced in a historical context. The problems that arise are then solved with the help of the basic theorems of spherical geometry, which are proved, for the most part, by adapting a historical proof. Van Brummelen declares early on that he appreciates the subject for its beauty; he finds the theorems elegant and often surprising; the proofs he provides have a visual impact that makes them informative and convincing. Chapter 1 addresses the question of how to measure the earth s radius, which is accomplished by ascending a mountain in the fashion of al-bīrūnī (around AD 1000), as opposed to the more familiar method of Eratosthenes (3rd century BC), which appears as an exercise. To complete the calculation one needs a sine table, which is constructed using ideas from Ptolemy (nd century AD). The chapter ends with a calculation of the distance to the moon (also following Ptolemy). Chapter provides an introduction to spherical geometry motivated by discussing the celestial sphere with its great circles representing the equator, ecliptic, and horizon. This leads to the question of how to describe the position of heavenly objects, the sun in particular, with respect to the celestial equator. Three answers are provided in the subsequent four chapters along with solutions to several other important problems: the ancient approach based on the spherical version of Menelaus s theorem, the medieval approach based on the rule of four quantities, and the modern approach based on Napier s discovery of how the parts of a spherical triangle are related. Crux Mathematicorum, Vol. 38(10), December 01

17 BOOK REVIEWS / 413 These first six chapters form the basis for the courses and workshops that the author has taught. The text seems to be suitable for a university course for math and science majors, or for liberal arts students who are not afraid of mathematical formulas and proofs. The final three chapters evolved from student projects; they deal with areas, angles, polyhedra, stereographic projection, and navigating by the stars. Each chapter ends with numerous problems that illustrate and extend the material; many are taken from historical textbooks but, unfortunately, the author does not provide the solutions. The book can be read with profit from cover to cover by the casual reader armed only with the basics of plane trigonometry very little is required beyond the geometric meaning of sine, cosine and tangent, and perhaps the sine and cosine laws. For the very casual reader whose interest is mainly in the pictures and anecdotes, the author has indicated the beginning and end of detailed arguments that, he says, can be omitted without losing the general flow of ideas. On the other hand, I found the mathematics fascinating; although I already knew much of the story I learned something, historical and mathematical, from every chapter. Mathematical Excursions to the World s Great Buildings by Alexander J. Hahn Princeton University Press, 01 ISBN: , Hardcover, ix pages, US$49.50 Reviewed by David Butt, architect and J. Chris Fisher, mathematician, Regina, SK By 196 the Italian city of Florence was already becoming prosperous and important, and so it required a cathedral commensurate with its new status. Of course the town fathers wanted the latest Gothic vaulting, but to make their structure exceptional the plans called for the largest dome imaginable. As was customary in those days, construction of the nave and transept went on for over a century with no idea of how, or even if it might be possible to build such a dome the octagonal drum on which the dome was to rest was 145 feet across, and it had to support a weight that was perhaps twenty times greater than that of any dome that had ever been built, yet obtrusive exterior buttresses were forbidden by the commission that oversaw the construction. Moreover, the drum itself reached 180 feet above the floor, a height which made infeasible the timber bracing built up from the ground that traditionally supported a dome as it was being constructed. An innovative solution was needed, and Filippo Brunelleschi ( ), a brilliant mathematician, artist, craftsman, inventor, and architect had the skill to devise that solution. Alexander Hahn lays a readable foundation for the breakthrough building projects we witness today. His analysis of historical structures should make every student of architecture proud. What better record do we have of man s creative evolution than our architecture? Architects and master builders rely heavily on the creative engineering skills of their design partners. The story of the building of Copyright c Canadian Mathematical Society, 013

18 414/ BOOK REVIEWS the Florence Cathedral is one of a half-dozen compelling excursions to the world s great buildings. Perhaps even more exciting was the construction of the Sydney Opera House ( ), which also required its designers to devise innovative solutions to numerous unprecedented problems; then, at the last minute, they had to bring in a bit of elementary mathematics to avert inundation by cost overruns. The author has organized his book around two historical narratives: one explores the architectural form and structure of a sampling of great buildings; the other develops mathematics from a historical perspective. In addition to six stories that are recounted in some detail, there are numerous brief discussions, some quite interesting, but many only superficial. All, however, are accompanied by well-chosen pictures and diagrams. As for the mathematics in the book, the most successful deal with analysis of thrusts, loads, tensions, and compressions. Hahn uses modern mathematics (vectors, trigonometry, and calculus) to analyze structures whose builders had no access to such tools; the builders relied instead on experience, ingenuity, and faith. For example, the author is able to explain why cracks developed in Brunelleschi s magnificent dome, and to assure us that corrective measures are in place to secure the stability of the structure for centuries to come. On the other hand, we suspect that readers of this journal would find most of the mathematics in the book to be boring; much of it falls far short of providing insight into the architecture. Large chunks of the book are devoted to elementary mathematics that has only a tenuous relationship to the main topic. It is not clear to us who the intended audience for those passages might be perhaps the author used the book for a mathematics course that he taught, so he felt obliged to include more mathematics. He does not say. Architectural students would likely skip the mathematical digressions because they are distracting; readers with little background in mathematics would probably find the content too concise and not particularly well explained. In the chapter on Renaissance buildings, for example, the fifteen pages devoted to perspective drawings consist of calculations involving Cartesian coordinates that not only explain very little, but they are most certainly inappropriate for the job. After a messy three-page calculation to demonstrate that a circle becomes an ellipse in a perspective drawing, in place of the steps that determine the endpoint of the major axis he declares that, This is an unpleasant computation that we will omit. After all that effort, we fail to learn how a Renaissance architect might have used drawings to design his building projects. It might be unusual for a review appearing in a mathematics journal to advise the reader to skip the mathematics, but following that advice, you will likely find Hahn s book fascinating reading. Crux Mathematicorum, Vol. 38(10), December 01

19 MURRAY S. KLAMKIN / 415 PROBLEM SOLVER S TOOLKIT No. 3 Murray S. Klamkin The Problem Solver s Toolkit is a new feature in Crux Mathematicorum. It will contain short articles on topics of interest to problem solvers at all levels. Occasionally, these pieces will span several issues. On a Triangle Inequality [Ed. : Note, this article originally appeared in Crux Mathematicorum Volume 10, No. 5 [1984 : ].] It is well known [4, p. 18] that, if A, B, C are the angles of a triangle, then sin A + sin B + sin C 3 3, (1) with equality if and only if A = B = C. (A proof is immediate from the concavity of the sine function on the interval [0, π].) Vasić [1] generalized (1) to 3 yz x sin A + y sin B + z sin C x + zx y + xy, () z where x, y, z > 0. In [], the author showed that () was a special case of the two-triangle inequality 4(xx sin A sin A + yy sin B sin B + zz sin C sin C ) yz x + zx y + xy z Here we strengthen () to y z x + z x y + x y x sin A + y sin B + z sin C 1 (yz + zx + xy) r x + y + z xyz z. (3). (4) We start with the polar moment of inertia inequality [3], (w 1 + w + w 3 )(w 1 R 1 + w R + w 3 R 3) w w 3 α 1 + w 3 w 1 α + w 1 w α 3, in which w 1, w, w 3 are arbitrary nonnegative numbers; α 1, α, α 3 are the sides of a triangle A 1 A A 3 ; and R 1, R, R 3 are the distances from an arbitrary point to Copyright c Canadian Mathematical Society, 013

20 416/ PROBLEM SOLVER S TOOLKIT the vertices of the triangle. Taking R 1 = R = R 3 = R, the circumradius of the triangle, and using the power mean inequality, we obtain w w 3 α 1 + w 3 w 1 α + w 1 w α 3 w w 3 + w 3 w 1 + w 1 w w w 3 α 1 + w 3 w 1 α + w 1 w α 3 w w 3 + w 3 w 1 + w 1 w ª, R(w 1 + w + w 3 ) w w 3 + w 3 w 1 + w 1 w w w 3 α 1 + w 3 w 1 α + w 1 w α 3. (5) Now letting w 1 = yz x, w = zx y, w 3 = xy z, and using α i = R sin A i in (5), we obtain (4). There is equality if and only if α 1 = α = α 3 and the centroid of the weights w 1, w, w 3 at the respective vertices of the triangle coincides with the circumcentre. This entails that or, equivalently, that w 1 = w = w 3, x = y = z. We have therefore shown that equality holds in (4) if and only if the triangle is equilateral. Finally, to show that (4) is stronger than (), we must establish that yz r 3 x + zx y + xy x + y + z (yz + zx + xy) z xyz or, equivalently, that 3(y z + z x + x y ) xyz(x + y + z)(yz + zx + xy). Letting x = 1 u, y = 1 v, and z = 1 w shows that this is equivalent to 3(u + v + w ) (u + v + w) (vw + wu + uv). Since u + v + w u + v + w 3 3 by the power mean inequality, it suffices finally to show that u + v + w vw + wu + uv, 3 3 and this is equivalent to 4 (v w) + (w u) + (u v) 0. Crux Mathematicorum, Vol. 38(10), December 01

21 MURRAY S. KLAMKIN / 417 References [1] P. M. Vasić, Some Inequalities for the Triangle, Publ. Eletrotehn. Fak. Univ. Beogradu, Ser. Mat. Fiz., No No. 301 (1964) [] M. S. Klamkin, An Identity for Simplexes and Related Inequalties, Simon Stevin, 48 ( ) [3] M. S. Klamkin, Geometric Inequalities via the Polar Moment of Inertia, Mathematics Magazine, 48 (1975) [4] O. Bottema et al., Geometric Inequalities, Wolters-Noordhoff Publ., Groningen, A Taste Of Mathematics Aime-T-On les Mathématiques ATOM One square is deleted from a square checkerboard with n squares. Show that the remaining n 1 squares can always be tiled with shapes of the form which cover three squares. This appears as problem 6 from Mathematical Olympiads Correspondence Program ( ) by Edward J. Barbeau which is Volume I of the Canadian Mathematical Society s booklet series ATOM. Booklets in the ATOM series are designed as enrichment materials for high school students with an interest in and aptitude for mathematics. Some booklets in the series will also cover the materials useful for mathematical competitions at national and international levels. There are currently 13 booklets in the series. For information on tiles in this series and how to order, visit the ATOM page on the CMS website: Copyright c Canadian Mathematical Society, 013

22 418/ Triangles Whose Angles Satisfy B = 90 + C RECURRING CRUX CONFIGURATIONS 9 J. Chris Fisher Triangles Whose Angles Satisfy B = 90 + C The main property of the triangles we look at this month first appeared in Crux in 000: Problem 55 (April) [000: 177; 001: 70-71] (proposed by Antreas P. Hatzipolakis and Paul Yiu; reworded here). In a triangle ABC with B > C, B = 90 + C if and only if the centre of the nine-point circle lies on the line BC. The proposers also called for the reader to show that if, in addition, A = 60, then the 9-point centre N is the point where the bisector of angle A intersects BC. Of course, as we saw in Part 3 of this column, AN bisects angle A whenever A = 60, whatever B and C might be. Problem 765 [00: 397; 003: ] (Proposed by K.R.S. Sastry). Derive a set of side-length expressions for the family of Heron triangles ABC in which the nine-point centre N lies on the line BC. Michel Bataille s solution invoked Problem 55; specifically, he proved that ABC has integer sides and area while B = 90 + C if and only if there exists a primitive Pythagorean triple (m, n, k) with m > n and a positive integer d for which a = d(m n ), b = dkm, c = dkn. The area, given by Heron s fromula, equals d mn(m n )/, which is an integer since m and n have opposite parity. The original problem required further that N lie in the interior of the segment BC; this will be the case if m > 3n. For an explicit example having B between N and C we can set m = 4, n = 3, k = 5, and d = 1; the resulting triangle has a = 7, b = 0, c = 15, sin B = 4/5, sin C = 3/5, and area = 4. As part of his solution, Sastry added a list of nine properties that are equivalent for any triangle ABC: (1) The nine-point centre is on BC. () B C = 90. (3) tan B tan C = 1. (4) OA BC (where O is the circumcentre of ABC). (5) AH is tangent to the circumcircle of ABC (where H is the orthocentre). Crux Mathematicorum, Vol. 38(10), December 01

23 J. CHRIS FISHER / 419 (6) AH is tangent to the nine-point circle of ABC. (7) BC bisects segment AH. (8) AN = 1 OH. (9) AC and BC trisect OCH. Problem 867 [003: 399; 004: ] (proposed by Antreas P. Hatzipolakis and Paul Yiu). With vertices B and C of triangle ABC fixed and its nine-point centre sliding along the line BC, the locus of the vertex A is the rectangular hyperbola (excluding B and C) whose major axis is BC. When B = C + 90 every point except B on the branch of the hyperbola through B is the vertex of such a triangle; when C = B + 90, A C sweeps out the branch through C. Christopher J. Bradley remarked that this is the hyperbolic analogue of the familiar theorem about angles inscribed in semicircles: When BC is the diameter of a semicircle containing A, then B + C = 90 ; when BC is the major axis of a rectangular hyperbola containing A, then B C = 90. I conclude this month s essay, as well as this series, by recalling an observation made by Charles W. Trigg [1978: 79] that was included in Part of the series: If the sides of ABC satisfy c + a = b while the angles satisfy A = C + 90, then the sides a, b, c are in the ratio ( 7 + 1) : 7 : 7 1. Call for Problem of the Month Articles We have introduced a few new features this volume. One of these new features is the Problem of the Month which is dedicated to the memory of former CRUX with MAYHEM Editor-in-Chief Jim Totten. The Problem of the Month features a problem and solution that we know Jim would have liked. Do you have a favourite problem that you would like to share? Write it up and send it to the editor at crux-editors@cms.math.ca. Articles featured in the Problem of the Month should be instructive, anecdotal and entertaining. We are also seeking an editor for this column. If you are interested, or have someone to recommend, please contact the editor at the address above. Copyright c Canadian Mathematical Society, 013

24 40/ PROBLEMS PROBLEMS Toutes solutions aux problèmes dans ce numéro doivent nous parvenir au plus tard le 1 juin 014. Une étoile ( ) après le numéro indique que le problème a été soumis sans solution. Chaque problème sera publié dans les deux langues officielles du Canada (anglais et français). Dans les numéros 1, 3, 5, 7, et 9, l anglais précédera le français, et dans les numéros, 4, 6, 8, et 10, le français précédera l anglais. Dans la section des solutions, le problème sera publié dans la langue de la principale solution présentée. La rédaction souhaite remercier Jean-Marc Terrier, de l Université de Montréal, et Rolland Gaudet, de Université de Saint-Boniface, Winnipeg, MB, d avoir traduit les problèmes Proposé par John Lander Leonard, Université de l Arizona, Tucson, AZ. Les nombres de 0 à 1 sont répartis au hazard autour d un cercle. (a) Montrer qu il doit exister un groupe de trois nombres adjacents dont la somme vaut au moins 18. (b) Trouver le n maximal tel qu il doit exister un groupe de trois nombres adjacents dont la somme vaut au moins n Proposé par Marcel Chiriţă, Bucharest, Romania. Résoudre le système suivant for x, y R. x + 6 = 1 3 x + 3 y = Proposé par George Apostolopoulos, Messolonghi, Grèce. Soit a, b et c trois nombres réels positifs tels que a + b + c = Trouver la valeur maximale de l expression a + b + b + c + c + a Proposé par Václav Konečný, Big Rapids, MI, É-U. Soit ABC un triangle acutangle inscrit dans un cercle Γ, et soit A et B les pieds des hauteurs respectives issues de A et B. Soit respectivement P A et P B les deuxièmes intersections des cercles de diamètre BA et CB avec Γ. Déterminer l angle entre les droites AP A et BP B. Crux Mathematicorum, Vol. 38(10), December 01

25 PROBLEMS / Proposé par José Luis Díaz-Barrero, Université Polytechnique de Catalogne, Barcelone, Espagne. Soit a, b et c les longueurs des côtés d un triangle ABC de hauteurs h a, h b, h c et de rayon circonscrit R. Montrer que 1 a + 1 b + 1 c R + h a + h b + h c 6 > Proposé par Michel Bataille, Rouen, France. Montrer que lim ny n k=1 (k) ( k 1) n (k + 1) k) = 1. (n Proposé par Panagiote Ligouras, École Secondaire Léonard de Vinci, Noci, Italie. Soit m a, m b et m c les médianes et k a, k b et k c les symédianes d un triangle ABC. Si n est un entier positif, montrer que ma k a n + mb k b n + mc k c n Proposé par Albert Stadler, Herrliberg, Suisse. Soit n un entier non négatif. Montrer que X Z k n k e t t k+1 dt k! k=0 1 = nx k=0 S(n, k) k +, où l on pose k n = 1 pour k = n = 0 et S(n, k) sont les nombres de Stirling de seconde espèce, définis par la récurrence S(n, m) = S(n 1, m 1) + ms(n 1, m), S(n, 0) = δ 0,n, S(n, n) = 1, pour n, m 0. A noter que S(n, m) = 0 quand m > n. [Ed : A noter, ceci est la version corrigée du problème 3687.] Proposé par Constantin Mateescu, Zinca Golescu Collège National, Pitesti, Roumanie. Soit respectivement R, r et s le rayon du cercle circonscrit, celui du cercle inscrit et le semi-périmètre d un triangle ABC pour lequel on dénote K = Montrer que s = 4R(K 1) R(K + 1) + r. X cyclique sin A. Copyright c Canadian Mathematical Society, 013

26 4/ PROBLEMS Proposé par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie. Soit n un entier. Calculer Z Z e x e y n dxdy. 0 0 x y Proposed by John Lander Leonard, University of Arizona, Tucson, AZ. The numbers 0 through 1 are randomly arranged around a circle. (a) Show that there must exist a trio of three adjacent numbers which sum to at least 18. (b) Determine the maximum n such that there must exist a trio of three adjacent numbers which sum to at least n Proposed by Marcel Chiriţă, Bucharest, Romania. Solve the following system for x, y R. x + 6 = 1 3 x + 3 y = Proposed by George Apostolopoulos, Messolonghi, Greece. Let a, b, and c be positive real numbers such that a + b + c = Find the maximum value of the expression a + b + b + c + c + a Proposed by Václav Konečný, Big Rapids, MI, USA. Let an acute triangle ABC be inscribed in the circle Γ, and A and B be the feet of the altitudes from A and B respectively. Let the circle on diameter BA intersect Γ again at P A, and the circle on diameter CB intersect Γ again at P B. Determine the angle between the lines AP A and BP B Proposed by José Luis Díaz-Barrero, Universitat Politècnica de Catalunya, Barcelona, Spain. Let a, b, c be the lengths of the sides of a triangle ABC with altitudes h a, h b, h c and circumradius R. Prove that 1 a + 1 b + 1 c R + h a + h b + h c 6 > 3. Crux Mathematicorum, Vol. 38(10), December 01